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kupik [55]
3 years ago
7

WIL GIVE BRAINLISEST PLEASE HELP SPEED ACCELERATION GRAPHS LABEL THE GRAPHS PLEASEEeeee

Physics
1 answer:
Varvara68 [4.7K]3 years ago
4 0

Answer:

Take the numbers and fill them in graph.

Plot time as X and speed as Y and afterward join then like a line graph.

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*How much energy is<br>transferred in lifting a 5 kg<br>Mass 3m​
AlexFokin [52]

Answer:

147 J

Explanation:

The energy transferred to potential energy is :

U = m * g * h = (5 kg) * (9.8 m/s^2) * (3 m) = 147 J

3 0
3 years ago
If a car travels 60 mph for a distance of 180 miles, how much time<br> did it take?
jolli1 [7]

Answer:

3 hours

Explanation:

180 divided by 60 (mph means miles per hours by the way)

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3 years ago
Hlo <br><br> what is a force........??
Paha777 [63]

Explanation:

In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity, i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity.

Formula

Newton's Second Law

F = m * a

F = force

m = mass of an object

a = acceleration

5 0
3 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
4 years ago
Draw the following patterns as they would appear when viewed through a compound microscope.:
goblinko [34]
2 j x b
H c e c hope this helppppsssssssss
3 0
3 years ago
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